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Special Relativity: Energy and Mass

  1. Jul 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Two parts to the problem: 1.) Is the following a good argument from special relativity for the equivilence of energy and mass?

    2.) If 1.) is true, why couldn't a similar argument have been made with pre-Einstein physics?



    2. Relevant equations

    Just two:

    E = mc2/(1 - (v/c)2)1/2

    E = (1/2)mv2



    3. The attempt at a solution

    Okay, starting with Einstein's energy equation, then assuming some energy is added to the moving particle. It seems pretty obvious that energy is convertible into mass (because of the speed of light squared factor maybe? see below). But I have some questions. First, of course, does the following show that energy is interchangeable with mass:


    E = mc2/(1 - (v/c)2)1/2

    Particle absorbs some energy E0:

    E2 = (mc2 + E0)/(1 - (v/c)2)1/2

    Factor out c^2, which gives:

    E2 = (m + E0/c2)c2/(1 - (v/c)2)1/2


    But an energy divided by a velocity squared leaves units of mass (kilogram meters^2/seconds^2, divided by meters^2/seconds^2 ).
    Looking at the equation then, we could say that m + E0/c^2 equals a new mass, M, so that

    m + E0/c2 = M Which can be written in terms of Einstein's energy equation:

    E2 = Mc2/(1 - (v/c)2)1/2


    Which seems to indicate that adding energy E0 to E yields an increase in mass.


    Is that fairly decent so far? If so, here's my problem:




    If the above is right, why can't we make the same argument for Newtonian physics? (or maybe, why didn't they?)

    Starting with the kinetic energy equation:



    E = (1/2)mv2

    add E0

    E2 = (1/2)mv2 + E0

    factor out (1/2)v2, which gives:


    E2 = (m + E0/v2) *(1/2)v2

    where E0/v2 is again in units of mass, because an energy divided by a squared velocity leaves units of mass. This seems to me to indicate that even in pre-Einstein physics energy is interchangeable with mass. Clearly this is incorrect (and yes, it is purely academic, since obviously Newton's energy equations are less accurate than Einstein's).


    So, can someone please explain to me what is wrong with all this? Was I correct in the first part (relativity part)?


    Thanks, as always!
     
  2. jcsd
  3. Jul 31, 2008 #2

    tiny-tim

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    Hi Battlemage!! :smile:

    (have a square-root: √ :smile:)
    I don't understand either the statement "Particle absorbs some energy E0", or the reason for putting E0 where it is in the last equation. :confused:
     
  4. Jul 31, 2008 #3
    Okay. What I mean is that a moving particle is, say, hit with some sort of sub-atomic particle that has an energy of E0 and then that energy is added to the moving mass when the sub-atomic particle becomes a part of the mass. Just like if you "add energy" to some ice it becomes more energetic and turns into water, but of course we're talking about some nuclear reaction.

    So, in the closed system, the particle has some energy, as defined by Einstein's equation, and then more energy is added to it in the form of E0 (whether it be in terms of heat, some sub-atomic particle, etc).



    The E0 is added into the equation because if you take an energy and add another to it, you get a sum of the total energy, do you not? (regardless of how the energies actually add up- I wonder how they do, since adding velocities doesn't work the same way in a Lorentz invariant model that it does in a Galilean invariant model)



    Does that help to clarify? If I am way off base here, by all means, show me how I would take a particle in special relativity and add some amount of energy to it.



    (by the way, the use of the notation E0, added to E = mc2/√(1 - (v/c)2), comes from Einstein's laymen book on Relativity. He used the notation himself, but didn't really get into the details [which of course is why I'm asking the question], so I gotta believe that at least a little of the original post is right, or at least on the right track. Course, I could be way off and totally misunderstood it)



    Thanks for the √, by the way. :)
     
  5. Mar 2, 2010 #4
    How did Einstein come up with the Energy equation E = m(c^2)/√[(1 - (v/c)^2] without the basic assumption that:

    m = m_0/√[1 - (v/c)^2]. m = mass in motion, m_0 = mass at rest. Also known as the relativistic mass equation.

    In the above equation on the top line of this post, the "m" is equivalent to the "m_0" in the equation on line 4.
    This relativistic mass equation was taught to us in high school physics and in physics texts, it is taught as Gospel as if it came from God. However, I have never seen a proof that this really is true (although I believe it is.)

    If, on the other hand, there is a way to dervive that Energy equation E = m(c^2)/√[(1 - (v/c)^2] directly without using the relativistic mass equation on line 4 (the "God-given equation"] as a basis, then this Energy equation "proves" the relativistic mass equation.
     
  6. Apr 8, 2010 #5

    You can derive the relativistic kinetic energy equation (assuming the particle doesn't cause heat through friction by moving) by integrating, from 0 to v, mγ3v dv, with m being constant for low velocities.

    γ is the Lorentz factor, which you can derive by assuming that the speed of light is constant for all inertial frames and the laws of physics are the same in all inertial frames.

    the result of this integration is:

    mc2 (1/√(1-(v/c)2) - 1 )





    This link does the antiderivative of that (i.e. no bounds on the integral- but the dummy variable is x there, rather than v)

    http://integrals.wolfram.com/index.jsp?expr=m*(1/sqrt(1+-+(x/c)^2+)^3+*+x&random=false

    If you replace x with v and then subtract from that the same antiderivative but replace x with zero you get the above result. Of course I also factored out the term mc2 and wrote it in the order m, then c2, while Wolfram didn't do either.)




    As far as the energy equation from momentum, you can "derive" this by trial and error, when you see that if you use the Lorentz transformation for transforming bodies, you will find that momentum isn't conserved UNLESS you multiply the classical definition of momentum by γ.


    Assume two bodies collide perfectly inelastically, and pick one of them to have an initial velocity of 0 with respect to one frame. The only way P1 = P2 while using the Lorentz transformation is if P1 and P2 are each defined in the relativistic way.

    I haven't seen how Einstein actually derived it, but it would make perfect sense to try multiplying by the Lorentz factor, since the transformation is linear and multiplying by a scaler is a linear operation, and because (most obviously) you would be working in a space-time in which the Lorentz transformation is the law by which coordinate transformations are made. So, if you were trying to "guess" the correct formula, simply multiplying momentum by the Lorentz factor would be at or near the top of the list of "things to try."
     
    Last edited: Apr 8, 2010
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